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As we said before, error correction is much more difficult than error detection In error detection, the receiver needs to know only that the received codeword is invalid; in error correction the receiver needs to find (or guess) the original codeword sent We can say that we need more redundant bits for error correction than for error detection Figure 107 shows the role of block coding in error correction We can see that the idea is the same as error detection but the checker functions are much more complex
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Figure 107 Structure of encoder and decoder in error correction
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I Checker I
Unreliable transmission
Codeword
In bits
Example 103
Let us add more redundant bits to Example 102 to see if the receiver can correct an error without knowing what was actually sent We add 3 redundant bits to the 2-bit dataword to make 5-bit codewords Again, later we will show how we chose the redundant bits For the moment let us concentrate on the error correction concept Table 102 shows the datawords and codewords Assume the dataword is 01 The sender consults the table (or uses an algorithm) to create the codeword 01011 The codeword is corrupted during transmission, and 01001 is received (error in the second bit from the right) First, the receiver finds that the received codeword is not in the table This means an error has occurred (Detection must come before correction) The receiver, assuming that there is only 1 bit corrupted, uses the following strategy to guess the correct dataword
ERROR DETECTION AND CORRECTION
Table 102 A code for error correction (Example 103)
Dataword Codeword
00 01 10
00000 01011 10101 11110
I Comparing the received codeword with the first codeword in the table (01001 versus 00000), the receiver decides that the first codeword is not the one that was sent because there are two different bits 2 By the same reasoning, the original codeword cannot be the third or fourth one in the table 3 The original codeword must be the second one in the table because this is the only one that differs from the received codeword by 1 bit The receiver replaces 01001 with 01011 and consults the table to find the dataword 01
Hamming Distance
One of the central concepts in coding for error control is the idea of the Hamming distance The Hamming distance between two words (of the same size) is the number of differences between the corresponding bits We show the Hamming distance between two words x and y as d(x, y) The Hamming distance can easily be found if wc apply the XOR operation (ffi) on the two words and count the number of Is in the result Note that the Hamming distance is a value greater than zero
The Hamming distance between two words is the number of differences between corresponding bits
Example 104
Let us find the Hamming distance between two pairs of words 1 The Hamming distance d(OOO, 011) is 2 because 000 ffi 011 is 011 (two Is) 2 The Hamming distance d(10101, 11110) is 3 because 10101 ffi 11110 is 01011 (three Is)
Minimum Hamming Distance
Although the concept of the Hamming distance is the central point in dealing with error detection and correction codes, the measurement that is used for designing a code is the minimum Hamming distance In a set of words, the minimum Hamming distance is the smallest Hamming distance between all possible pairs We use d min to define the minimum Hamming distance in a coding scheme To find this value, we find the Hamming distances between all words and select the smallest one
The minimum Hamming distance is the smallest Hamming distance between all possible pairs in a set of words
SECTION iO2
BLOCK CODiNG
Example 105
Find the minimum Hamming distance of the coding scheme in Table 101
Solution
We first find all Hamming distances
d(OOO, 011) = 2 d(Oll,110)=2 d(OOO, 101) = 2 d(W1, 110) = 2 d(OaO, 110) = 2 d(Oll, 101)
The dmin in this case is 2
Example 106
Find the minimum Hamming distance of the coding scheme in Table 102
Solution
We first find all the Hamming distances d(OOOOO, 01011) = 3 d(01011, 10101) =4 The dmin in this case is 3 d(OOOOO, 10101) 3 d(OlO11, 11110) = 3
d(OOOOO, 11110) = 4 d(10101, 11110) =3
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